Telescoping series: $\sum\limits_{n=1}^{∞}[\tan^{-1}(2n+1)-\tan^{-1}(2n-1)]$

Question

In this question that was asked today the OP wrote that $$\begin{align}\sum\limits_{n=1}^{∞}[\arctan(2n+1)-\arctan(2n-1)]&=\arctan\infty-\arctan 1\\ &=\frac{\pi}{2}-\frac{\pi}{4}\\&=\frac{\pi}{4} \end{align}$$

I don't really understand why there is a $\arctan\infty$ term. Surely every single term other than $-\arctan 1$ has been cancelled out? Now I know that I am wrong, as obviously the value of the summation cannot be negative, but I'm not sure where I am wrong.

I have thought of considering the similar finite series, $$\sum\limits_{n=1}^k[\arctan(2n+1)-\arctan(2n-1)]=\arctan(2k+1)-\arctan1$$ and letting $k\to\infty$, so that $$\lim_{k\to\infty}\sum\limits_{n=1}^k[\arctan(2n+1)-\arctan(2n-1)]=\arctan\infty-\arctan1$$ as required, but I still can't see why all the terms other than $-\arctan1$ don't cancel out, as the upper limit actually is $\infty$.

Thank you for your help.

Answer 1

A few intermediate steps might be helpful.

We obtain \begin{align*} \color{blue}{\sum_{n=1}^{\infty}}&\color{blue}{\left[\arctan(2n+1)-\arctan(2n-1)\right]}\\ &=\lim_{N\to\infty}{\sum_{n=1}^{N}\left[\arctan(2n+1)-\arctan(2n-1)\right]}\\ &=\lim_{N\to\infty}\left[\arctan(2N+1)-\arctan(1)\right]\tag{1}\\ &=\lim_{N\to\infty}\arctan(2N+1)-\lim_{N\to\infty}\arctan(1)\tag{2}\\ &=\frac{\pi}{2}-\frac{\pi}{4}\\ &\,\,\color{blue}{=\frac{\pi}{4}}\\ \end{align*}

Comment:

• In (1) we apply telescoping.

• In (2) we use $\lim_{N\to\infty}\arctan(N)=\frac{\pi}{2}$ and $\lim_{N\to\infty} a=a$.